POJ3279 Flip coin (state compression Dfs)

Source: Internet
Author: User
Tags first row

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The main idea: a n*m matrix flip coin, turn a coin will involve around four, ask all the coins face up to the minimum need to turn over which few coins.


It is not the same as how many times the previous question asked, this time to record the status.

First, we first think of enumerating the first row, and then to the other rows, as long as the top of a coin is upside down, and finally check.

The first place to learn is how to enumerate the first row of the flipping method:

for (int i=0;i< (1<<m); i++)
	 dfs (i);
for (int j=0;j<m;j++)
		if ((i>>j) &1) {
			flip (0,j);

It was marvelous, and all at once a 2^m of the whole arrangement.

As for the status record, you can copy the array A to the array temp each time, and then open an array vis, if you turn over a piece, vis[x][y] becomes 1, and when the final check is satisfied, the VIS value is copied to the other array res, because the VIS array is cleared 0 each time, So we have two sets of solutions to compare the size of the solution can be used in the RES array to record the best answer.

At the end of the output, if ANS does not change, it shows no solution, output impossible;

Otherwise, the output res array.

The following code is attached:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <
String> #include <cmath> using namespace std;
const int MAXN = 18;
int m,n,ans=1e6,dir[4][2]={1,0,-1,0,0,1,0,-1};

	void Flip (int x,int y) {vis[x][y] = 1;
	Temp[x][y] =!temp[x][y];
	if (x-1>=0) temp[x-1][y] =!temp[x-1][y];
	if (x+1<n) temp[x+1][y] =!temp[x+1][y];
	if (y-1>=0) temp[x][y-1] =!temp[x][y-1];
if (y+1<m) temp[x][y+1] =!temp[x][y+1];
	} bool Check () {for (Int. i=0;i<n;i++) for (int j=0;j<m;j++) if (Temp[i][j]) return false;
return true;
	} void Dfs (int k) {memcpy (temp,a,sizeof (a));
	memset (vis,0,sizeof (VIS));
	int cnt = 0;
			for (int j=0;j<m;j++) {if ((k>>j) &1) {Flip (0,j);
	  	}} for (int i=1;i<n;i++) for (int j=0;j<m;j++) if (temp[i-1][j]==1) {flip (i,j);
		} if (check () &&cnt<ans) {ans = cnt;memcpy (res,vis,sizeof (VIS));
	}} int main () {scanf ("%d%d", &n,&m);
	for (int i=0;i<n;i++) for (int j=0;j<m;j++) scanf ("%d", &a[i][j]);
	for (int i=0;i< (1<<m); i++) Dfs (i);
	if (ans==1e6) cout<< "Impossible" <<endl;
		else for (int i=0;i<n;i++) {for (int j=0;j<m-1;j++) {cout<<res[i][j]<< "";
		} cout<<res[i][m-1];
} return 0; }

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